Power series of $1/(1+x^2)$ around arbitrary a I've been trying to show that $f(x)=1/(1+x^{2})$
  has a power series expansion around any point $a\in\mathbb{R}$
 .
If $a=0$
  then I can see that for $\left|x\right|<1$
  (s.t. $\left|x^{2}\right|<1$
 ) we have $1/(1+x^{2})=1/(1-(-x^{2}))=\sum(-x^{2})^{k}$.
Is there are similar strategy of simplifying to a geometric series for general $a\in\mathbb{R}$
 ? Or are there other smart ways of attacking the problem?
I would be very grateful for any answers or hints on how to proceed.
 A: First use partial fraction as

$$ \frac{1}{(1+x^2)} = \frac{1}{2}\frac{1}{x+i} -\frac{1}{2} \frac{1}{x-i}. $$

Then we find the Taylor series at the point $x=a$ as

$$\frac{1}{2}\frac{1}{x+i}= \frac{1}{2}\frac{1}{(x-a)+(a+i)}= \frac{1}{2(a+i)\left(1+\frac{(x-a)}{a+i}\right)}=\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k(x-a)^k}{(a+i)^{k+1}}. $$

Do the other one and then try to simplify things and you will have a nice series representation.
A: Without going into the complex numbers...
Letting $u=x-a$, we have:
$$\frac{1}{1+x^2} = \frac1{1+a^2+2au+u^2}= \frac 1{a^2+1}\frac{1}{1+\frac{2au+u^2}{a^2+1}}$$
Then expand $\frac{1}{1+v}=\sum (-v)^k$ with $v=\frac{2au+u^2}{a^2+1}$.
That gives you a rather messy formula. The coefficient of $(x-a)^n$ is going to be gotten by finding the coefficient of $u^n$ in  $$\left(-\frac{2au+u^2}{1+a^2}\right)^i$$ for $\lceil n/2\rceil \leq i \leq n$ combine them. That's the $n-i$ coefficient of 
$$\left(-\frac{2a+u}{1+a^2}\right)^i$$
Which is $$\binom{i}{n-i}(2a)^{2i-n}\left(\frac{-1}{a^2+1}\right)^i$$
So the $n$th coefficient is (because we had the additional $\frac{1}{a^2+1}$ outside the sum:
$$\sum_{i=0}^n (-1)^i\frac{\binom{i}{n-i}(2a)^{2i-n}}{(a^2+1)^{i+1}} =\frac{1}{(1+a^2)^{n+1}}\sum_{i=0}^n (-1)^i\binom{i}{n-i}(2a)^{2i-n}(1+a^2)^{n-i}$$
Note that we can go from $0$ to $n$ because when $i<\lceil n/2\rceil$, $n-i>i$ and the coefficient evaluates to zero.
A: Since science gave a nice formal and general answer, for sure, the following is not a full answer to the question.
Applying the definitions (and being patient), you could  show that Taylor expansion around $x=a$ write as $$\frac{1}{1+x^2}=\sum_{n=0}^{\infty}\frac{(-1)^nP_n(a)}{(1+a^2)^{n+1}}(x-a)^n$$ with $$P_0(a)=1$$ $$P_1(a)=2a$$  $$P_2(a)=3a^2-1$$ $$P_3(a)=4a^3-4a$$ $$P_4(a)=5a^4-10a^2+1$$ $$P_5(a)=6 a^5-20 a^3+6 a$$ $$P_6(a)=7 a^6-35 a^4+21 a^2-1$$ in which some patterns seem to appear (different for odd and even values of $n$).
A: Here is my solution to this problem I came up with
$$
\begin{align*}
 \frac{1}{1+z^2}
 &= \frac{1}{(z-\mathrm i)(z+\mathrm i)} \\
 &= \frac{1}{2\mathrm i(z-\mathrm i)} - \frac{1}{2\mathrm i(z+\mathrm i)} \\
 &= \frac{1}{2\mathrm i}\left( \sum_{n=0}^\infty \frac{(-1)^n(z-x)^n}{(z-\mathrm i)^{n+1}} - \sum_{n=0}^\infty \frac{(-1)^n(z-x)^n}{(z+\mathrm i)^{n+1}} \right)\\
 &= \sum_{n=0}^\infty \frac{1}{2\mathrm i} \left( \frac{1}{(z-\mathrm i)^{n+1}} - \frac{1}{(z+\mathrm i)^{n+1}} \right)(x-z)^n.
\end{align*}
$$
This series has the radius of convergence $|x-\mathrm i|=\sqrt{1+x^2}$.
